x-1分之x-2除以(x+1 -x-1分之3),其中x=根号3-2

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x-1分之x-2除以(x+1 -x-1分之3),其中x=根号3-2

x-1分之x-2除以(x+1 -x-1分之3),其中x=根号3-2
x-1分之x-2除以(x+1 -x-1分之3),其中x=根号3-2

x-1分之x-2除以(x+1 -x-1分之3),其中x=根号3-2
答:
[(x-2)/(x-1)]÷[x+1-3/(x-1)]
=[(x-2)/(x-1)]÷[(x²-1-3)/(x-1)]
=[(x-2)/(x-1)]×(x-1)/(x²-4)
=1/(x+2)
=1/(√3-2+2)
=√3/3

原式=(x-2)/(x-1)÷[(x+1)(x-1)-3]/(x-1)
=(x-2)/(x-1)÷(x²-4)/(x-1)
=(x-2)/(x-1)*(x-1)/(x+2)(x-2)
=1/(x+2)
=1/(√3-2+2)
=√3/3